Polar coordinates




Polar coordinates of a point P.

In the plane we choose a fixed point O, and we call it the pole.
Additionally we choose an axis x through the pole and call it the polar axis.
On that x-axis, there is just 1 vector E such that abs(E)=1.
The pole and the polar axis constitute the basis of the polar coordinate system.

Now, we take a point P.
On the line OP we choose an axis u.
The number t is a value of the angle from the x-axis to the u-axis.
The number r is such that P = r.U
The numbers r and t define unambiguous the point P.
We say that (r,t) is a pair of polar coordinates of P.

 
    
One point P has many pairs of polar coordinates. If (r,t) is a pair of polar coordinates, (r, t + 2.k.pi) is also a pair of polar coordinates and additionally (- r, t + (2.k+1).pi ) is a pair of polar coordinates too.
Of course, k is an integer.

Examples

 
    
Figure 1:
polar coordinates of P are (2 , 0.3) or (-2 , 3.44) or (2 , -5.98) ...

Figure 2:
polar coordinates of P are (-1.4 , 3.6) or (1.4 , 0.46 ) or (1.4 , -5.8) ...

Figure 3:
polar coordinates of P are (3 , 5.7) or (3 , -0.58) or (-3 , 2.56 ) ...

The polar coordinates of the pole O are by definition (0,t) with t perfectly arbitrary.

From polar to cartesian coordinates.

A polar coordinate system is given and point P has polar coordinates (r,t). We choose a y-axis through the pole O and perpendicular to the x-axis. So, we have cartesian axes x and y. Call (x,y) the cartesian coordinates of P.

According to the previous definition, the cartesian coordinates of U are (cos t, sin t).
Since P = r.U, the cartesian coordinates of p are (r.cos t, r.sin t).

The transformation formulas are x = r.cos t, y = r.sin t

Figure 1: the cartesian coordinates of P are (1.91, 0.59)
Figure 2: the cartesian coordinates of P are (1.25, 0.62)
Figure 3: the cartesian coordinates of P are (2.50, -1.65)

From cartesian to polar coordinates.

We start with a cartesian coordinate system. We choose O as pole and the x-axis as polar-axis.

The cartesian coordinates of a point P are (x,y).
Choose the u-axis such that r > 0. then

 

        P = r U => P2 = r2  U2  => x2  + y2 = r2

        r = sqrt(x2  + y2)

        Now, choose a  t-value  such that x = r.cos t and y = r.sin t
In that way we have a pair of polar coordinates (r,t) of P.
Starting with that pair, all pairs are
(r, t + 2.k.pi) and (- r, t + (2.k+1).pi )

Exercise: Transform the thee pairs of cartesian coordinates of P above back to polar coordinates.

Polar equation of a curve.

Consider a connection between the polar coordinates of a point and suppose, that connection can be expressed in the form F(r,t)=0 or maybe in the explicit form r = f(t).
Such equation is a polar equation of a curve.
With each solution (ro,to) of the polar equation, corresponds a point with polar coordinates (ro,to). Generally the equation has an infinity number of such solutions and so, we have an infinity number of points. The set of all these points is the curve of the equation.

Each point P of that curve has at least one pair of polar coordinates who satisfy the equation. Note that, in general, not all pairs of polar coordinates of P are solutions of the equation.

Note that one curve can have different polar equations.

Examples

 
    
Exercise 1:
On this site you can type the polar equation and see the corresponding curve. First you must choose for 'polar' instead of 'cartesian'. You can choose for single or for multiple graphs.
Look at the curves of
4*cos(t) gives a circle
4*cos(2*t) gives a four-blade curve
1/(1-cos(t)) gives a parabola
1/(1-0.8*cos(t)) gives an ellipse with a focus in O
(1 + cos(t)) gives a cardioid
 
    
Exercise 2:
Plot the four-blade curve r= 4 cos(2*t). Find the polar equation of the circle that is tangent to the tops of the leaves. Plot this circle. Exercise 3:
Given the curve with polar equation r = 2.tan(t).sin(t). Plot the curve.
Point A on the curve has a t-value = pi/4. B is the mirror image of A relative to the polar axis. Find the area of the triangle OAB.

Exercise 4:
Given the curve with polar equation r = 2.tan(t).sin(t).
Point A is on the curve and |OA| = 1/sqrt(3). 0 < t < pi/2.
Find t.

Direction of a curve in polar coordinates

 
    
Suppose a curve c has polar equation r = f(t).
At a variable point P of the curve, we draw a tangent line and on that line we denote the axis b in the direction of increasing t-values.
This defines the angle n and the angle a.
The tan(n) defines the direction of the curve c in point P. This tan(n) is somewhat similar to the notion of slope in cartesian coordinates.
We'll calculate tan(t).
 
        t + n = a

=>      n = a - t

                 tan(a) - tan(t)
=>      tan(n) = -----------------
                 1 + tan(a).tan(t)

Say the variable point P has cartesian coordinates (x,y).
Then we know that

                 dy                  y
        tan(a) = ---  and  tan(t) = ---
                 dx                  x
Thus,

                   dy   y
                   -- - -
                   dx   x       x dy - y dx
        tan(n) = ---------- = --------------
                     dy   y     x dx + y dy
                 1 + -- . -
                     dx   x

From x = r cos(t) and y = r sin(t)
we have
        dx = dr.cos(t) - r.sin(t) . dt
        dy = dr.sin(t) + r.cos(t) . dt
From this we calculate

        x dy - y dx = r2 dt


From    x2  + y2  = r2  we find

        2 x dx + 2 y dy = 2 r dr     or

         x dx +  y dy =  r dr

Now, we can simplify tan(n)

                 r2 dt         r
        tan(n) = -------  = ------- =
                  r dr       dr/dt

                    r
        tan(n) = -----
                    r'

The following formula gives the direction of the curve c at any point P. Here r = f(t) is the equation of the curve and r' stands for (dr/dt) = f'(t). From r and r' you can calculate the direction n (see figure above ) at each point.
 
             r
tan(n) = --------
             r'

Exercise 1:
We take the curve with polar equation r = 4 cos(t). It is a circle with radius 2. The circle contains the pole and has its center on the polar axis. Draw the circle. Choose point P on the circle with polar coordinates (2 , pi/3). Draw OP and the tangent line in point P. Draw the angle n as in the previous figure. Calculate the angle with the formula tan(n) = r/r' and verify your result on your figure.

Exercise 2:
We take the curve with polar equation r = 1/(1-cos(t)). It is a parabola with cartesian equation y2 = 2x+1. Draw the parabola. Choose point P on the parabola with polar coordinates (2 , pi/3). Draw OP and the tangent line in point P. Draw the angle n as in the previous figure. Calculate the angle with the formula tan(n) = r/r' and verify your result on your figure.

Isogonal Curves

We look for the curves such that the direction n (see figure above ) of that curve is a constant k in each point.

 
    cot(n) = r'/r = (ln(r))'
Now
    (ln(r))' = k  for all t.
First take k = 0. Then ln(r) = constant and r is constant. The curves are the circles with midpoint in the origin. Now, take k not 0. Then
 
    (ln(r))' = k

      dln(r)
<=>  ------- = k
       dt

<=>   ln(r) = k t + constant

                  we denote the constant as ln(m)

<=>    ln(r) = k t +  ln(m)

<=>     r = m ekt

These isogonal curves are called the logarithmic spirals or the Bernouilli spirals.

Exercise:
Give different values to the parameters m and k and plot some spirals. Let t vary so you can see a few turns.

Calculate k so that the constant angle is 60 degrees. (answer: k = 1/sqrt(3) )

Tangent line parallel to the polar-axis

In previous figure we see that
 
        the tangent line is parallel to the polar axis

<=>     t + n = k.pi

<=>     tan(t) = - tan(n)

                    r
<=>     tan(t) = - ---
                    r'
The following formula gives the t-values of all points of a curve r=f(t), where the tangent line is parallel to the polar-axis.
The value r' stands for (dr/dt) = f'(t). To calculate the t-values, you have to solve this trigonometric equation.
 
             r
tan(t) = - -----
             r'

Tangent line orthogonal to the polar-axis

In previous figure we see that
 
        the tangent line is orthogonal to the polar axis

<=>     t + n = pi/2 + k.pi

<=>     tan(t) = cot(n)

                    r'
<=>     tan(t) =   ---
                    r
The following formula gives the t-values of all points of a curve r=f(t), where the tangent line is orthogonal to the polar-axis.
The value r' stands for (dr/dt) = f'(t). To calculate the t-values, you have to solve this trigonometric equation.
 
           r'
tan(t) = -----
           r

Investigation of a curve ; an example

We take the curve with polar equation r = cos(t/2).
 
        

From a cartesian equation to a polar equation

Suppose a curve C has a cartesian equation F(x,y) = 0.
If we replace, without thinking, x by r.cos(t) an y by r.sin(t), then we have a polar equation F(r.cos(t), r.sin(t)) = 0 of a curve C'.
We'll show that C = C'.
Suppose a curve C has a cartesian equation F(x,y) = 0.

If we replace x by r.cos(t) and y by r.sin(t), then we have a polar equation F(r.cos(t),r.sin(t))=0 of the curve C.


Example 1:
We want to find a polar equation of the curve y2 + x2 - 4 x = 0.
We replace x by r.cos(t) and y by r.sin(t).
 
      r2 cos2(t) + r2 sin2(t) - 4r cos(t) = 0

<=>   r2 -  4r cos(t) = 0

<=>   ( r= 0  or  r - 4 cos(t) = 0 )
r= 0 is the equation of the pole itself. So, the pole belongs to the curve.
The second part r = 4 cos(t) contains the pole too, since t = pi/2 gives r = 0.
The graph will not change if we omit the part r = 0.
The full curve has polar equation r = 4 cos (t).

Example 2:
We want to find a polar equation of the curve

 
       y2 (1 + x) - x2 (1 - x) = 0

     We replace x by r.cos(t) and y by r.sin(t).

      r2 sin2(t) ( 1 + r cos(t)) - r2 cos2(t) (1 - r cos(t)) = 0

<=>   r2 ( sin2(t) - cos2(t) + r cos(t) sin2(t) + r cos3(t)) = 0

<=>   r = 0   or  - cos(2t) + r cos(t) = 0

<=>   r = 0   or   cos(2t) = r cos(t)

                     cos(2t)
<=>   r = 0   or r = --------
                      cos(t)
Since r = cos(2t)/cos(t) contains the pole, we omit the first part r=0.
r = cos(2t)/cos(t) is the requested polar equation.

Exercise: Show that the hyperbola x y = 2 has a polar equation r2 sin(2t) = 4.

From a polar equation to a cartesian equation

Suppose a curve C has a polar equation G(r,t) = 0.
Sometimes that equation can be transformed to an equation
F(r.cos(t), r.sin(t)) = 0.
In the same way as above you can show that F(x,y) = 0 is the cartesian equation of C.
The transformation from G(r,t) = 0 to F(r.cos(t), r.sin(t)) = 0 is often very difficult or impossible!
Suppose a curve C has a polar equation G(r,t) = 0.

If that equation can be transformed to an equation
F(r.cos(t), r.sin(t))=0 then F(x,y)=0 is the cartesian equation of C.


Example 1
Suppose a curve C has a polar equation r2 = 4 . We can transform this in
 
        r2  cos2 (t)  + r2  sin2 (t)  - 4 = 0  <=>  x2  + y2  = 4

Example 2
Suppose a curve C has a polar equation r = -3 sin(t).
For t = 0, we find r = 0.Thus, the pole is part of C and we don't add a point to C if we multiply both sides by r.
C has polar equation r2 = -3 r sin(t).
The cartesian equation of C is x2 +y2 + 3y = 0. C is a circle.

Example 3
Suppose a curve C has a polar equation r = 1 + 2.cos(t).
If we choose t such that cos(t) = 0.5, then r = 0. Thus, the pole is part of C and we don't add a point to C if we multiply both sides of r = 1 + 2.cos(t) by r.

C has a polar equation r2 = r + 2.r.cos(t) <=> r = r2 - 2.r.cos(t)

Now take the curve C' : r = -( r2 - 2.r.cos(t)) . We'll show that each point of C belongs to the curve C'.
Therefore we take an arbitrary point P(ro,to) on C. Since P is on C we have:

 
      ro = ro2 - 2.ro.cos(to)    (*)
But the same point P has polar coordinates ( -ro, to + pi ). So, we can write P(-ro,to + pi). We'll show that P is on C'.
 
   P( -ro, to + pi) is on  C'
<=>
   - ro = -( ro2 + 2.ro.cos(to + pi) )
<=>
    ro = ro2 - 2.ro.cos(to)

     and this is true because of (*)
In the same way we can show that each point of C' is on C. (exercise)

Hence, we can write

 
    C has a polar equation      r = (+1 or -1).( r2 - 2.r.cos(t))

<=> C has a polar equation r2  = (r2  - 2 r cos(t))2

<=> C has a  cartesian equation x2  + y2  = ( x2  + y2 -2 x )2

More examples of curves with polar equation

There is a site on the net where you can see the graph of all famous curves with the polar equation and the cartesian equivalent (if any).
 
        Cardioid
        Cissoid of Diocles
        Cochleoid
        Conchoid
        Trifolium
        Folium
        Folium of Descartes
        Hyperbolic Spiral
        Lemniscate of Bernoulli
        Right Strophoid
        etc...

Go and look at

Famous Curves Index
There you will find the names of the curve, its history and some of its associated curves.

Dot product in polar coordinates

Take a polar coordinate system and two vectors
 
        u(r1 , t1 ) and v(r2 , t2 )

In the corresponding  cartesian coordinate system the two  vectors
have cartesian coordinates
        u(x1 , y1 ) and v(x2 , y2 )

with
        x1  = r1  cos(t1 )   x2  = r2  cos(t2 )


        y1  = r1  sin(t1 )   y2  = r2  sin(t2 )


The dot product u.v

        = x1 x2  + y1 y2


        = r1  cos(t1 ) . r2  cos(t2 ) + r1  sin(t1 ).r2  sin(t2 )


        = r1 r2 (cos(t1 )cos(t2 ) + sin(t1)sin(t2 ))


        = r1 r2  cos(t1 - t2 )

 
The dot product of two vectors u (r1 , t1 ) and v (r2 , t2 )

 is     u . v = r1 r2  cos(t1 - t2 )


Exercise:
P and Q are variable points with (sin(t) , t) and (2, 2t) as respective polar coordinates, and t is a parameter different from zero.

Find the values of t such that the dot product of OP and OQ is equal to 2.
Afterwards, take such value of t and draw the vectors OP and OQ. Check geometrically whether the scalar product is 2.

A line and its polar equation

A line through the pole.

Since all the points of the line correspond with a constant value of the angle t, the equation of such a line is t = constant.
Example :
t = 0 is the line of the polar axis
t = pi/2 is the y-axis
t = 1
...

A line d not containing the pole.

Draw a line through the pole perpendicular to the given line d.
Say N(ro,to) is the intersection point.
 
        A point P(r,t) is on line d

<=>     PN is orthogonal to ON

<=>     PN . ON = 0

<=>     (N - P). N = 0

<=>     N.N - P.N = 0

<=>     ro.ro.cos(0) - r.ro.cos(t - to) = 0

Since ro and cos(t - to) are not 0
                ro
<=>     r = -----------
            cos(t - to)
The pole is not on a line d.
The line l, through the pole and perpendicular to the line d, intersects line d in point N(ro,to).
The equation of line d is
 
            ro
      r = -----------
          cos(t - to)

Exercise 1:
Points A, B and C have polar coordinates: (2 , pi/2) (2 , 0) and (2 , pi). Find the polar equations of the lines AC, AB en BC.
(Ans: r = sqrt(2)/cos(t-pi/4) ; r = sqrt(2)/sin(t-pi/4) ; t = 0 )

Exercise 2:

 
r = 12/( 6*sin(t) - 4 cos(t) ) is a polar equation of a curve c.
Find the cartesian equation of c.

A circle and its polar equation

A circle with the pole as center.

In that case the polar equation is obvious.
Examples :
r = 2
r = 5
r = -5 (Same circle as r = 5)
r = 0 The pole
...
A circle with the pole as center and radius R has a polar equation r = R

An arbitrary circle

Say C(ro,to) is the center and R is the radius.
 
        P(r,t) is on the circle

<=>     || CP ||  = R


<=>     || CP ||2  = R2


<=>     (P - C)2 = R2


<=>     P2  - 2 P C + C2  = R2


<=>     r2  - 2 r ro cos(t - to) + ro2  = R2
A circle, with C(ro,to) as center and R as radius, has has a polar equation
 
        r2  - 2 r ro cos(t - to) + ro2  = R2

Exercise:
Find the center and the radius of the circle r(r + 4 sin(t)) = 21.
Ans: C(-2,pi/2) R=5

A special circle with

Take the circle with center C(R,0) and R is the radius.
For the equation we take the previous formula with ro = R and to = 0.
The equation is
 

        r2  - 2 r R cos(t) = 0

<=>     r = 0  or r = 2R cos(t)


<=>     r = 2R cos(t)
        (since this curve already contains the pole for t = pi/2)
A circle, with C(R,0) as center and R as radius, has has a polar equation
 
                r = 2R cos(t)

A conic section and its polar equations

See Polar equation of a not degenerated conic section.

Common points of two curves

A problem with common points of two curves

Say curve l is a line with a polar equation t = 1.
Say curve c is a circle with a polar equation r = 2.

In cartesian coordinates we find the coordinates of the common points by solving the system of the two equations.
But if we solve the system of the polar equations we find only one point with polar coordinates (2,1).

Now we'll show

The cause of this problem

Say curve c has a polar equation F(r,t)=0.
Call V the set of all solutions of the equation F(r,t)=0.
Then curve c is the set of all points corresponding with the set V.
With each element of V there is one and only one point of curve c.

Similarly, Say curve c' has a polar equation F'(r,t)=0.
Call V' the set of all solutions of the equation F'(r,t)=0.
Then curve c' is the set of all points corresponding with the set V'.
With each element of V', there is one and only one point of curve c'.

With an element of the intersection of V and V' corresponds a common point of c and c', BUT it is possible that V contains a solution (ro,to) and that V' contains a different solution (r1,t1) and that both solutions correspond with a common point of c and c'. Then (ro,to) and (r1,t1) are different polar coordinates of that same point.

In the example above we have
(-2,1) is a solution of t = 1 and this gives a point of the line l.
(2,1+pi) is a solution of r = 2 and this gives a point of the circle c.
Although these solutions are different, the corresponding point is a common point of the line and the circle.

A solution of this problem. A special property (P).

There are polar equations with the following special property!
 
We say that an equation F(r,t)=0 of a curve c has property (P)

                if and only if

ALL polar coordinates of EACH point of c are solutions of F(r,t)=0
Corollary:

Say curve c has a polar equation F(r,t)=0 with the property (P), and curve c' has a polar equation F'(r,t)=0. Now the set of solutions of the system of the two polar equations contains the coordinates of ALL the common points of the two curves and the problem has disappeared!

If at least one of the two equations has the property (P), then we can find all intersection points of the two curves by solving the system of the two equations. (The pole sometimes escapes, see below)

Equations with property (P)

The pole is a special point

Since the pole has so many polar coordinates, we always have to investigate this case separately.

Example: We calculate the intersection points of the curves with equation

 
  r = 4 cos(t)  and  r = 4 sin(t)
Since the first equation has property (P), the intersection points are the solutions of the system
 
        r = 4 cos(t)
        r = 4 sin(t)
The coordinates of the pole are not a solution of that system but the pole IS an intersection point of the curves.

To calculate the common points of curve c with equation F(r,t)=0 and c' with equation F'(r,t)=0, it is sufficient to solve the system of the equations F(r,t)=0 and F'(r,t)=0 if at least one of the equations has the property (P). But even then you have to investigate separately if the pole is a common point.

Exercise:
Find the intersection points of the curve r = cot(2t) and the line through point (1,0) and perpendicular to the polar axis.

 
     

Exercise:
Find the intersection points of the circle r = 2 sin(t) and the cardioid r = 1+cos(t).

 
      

An extensive example.

Given:
 
                             5
    K has equation  r = ----------------------------
                         ( 1 - 2 sin(t) + 3 cos(t) )

                              2
    L has equation  r = ----------------------
                        ( 2 cos(t) + sin(t) )


Calculate
  • The intersection points of K and L in polar coordinates.
  • The equation of K and L in cartesian coordinates.
  • The intersection points of K and L in cartesian coordinates.

First we investigate the nature of the curves K and L.
 

        

Since the equation of L has the property (P), the intersection points of K and L are the solutions of the system
 
           5
  r = -----------------------------
       ( 1 - 2 sin(t) + 3 cos(t) )

            2
  r = ----------------------
      ( 2 cos(t) + sin(t) )
From these equations we have
 
        5                                   2
   --------------------------- =    ----------------------
    ( 1 - 2 sin(t) + 3 cos(t) )     ( 2 cos(t) + sin(t) )


<=>  ....

<=>  4 cos(t) + 9 sin(t) = 2
and with the method from here we solve this equation.
 
<=>  ....

<=>  t1 = 2.518876  and t2 = -0.2137328

The corresponding values of r are :

     r1 = -1.92058   and  r2 = 1.14785

To convert the equation of K to a cartesian equation, we appeal on the properties of Polar equation of a conic section.

The conic section K has polar equations

 
           5                                    -5
  r = --------------------------  and   r = ----------------------------
       ( 1 - 2 sin(t) + 3 cos(t) )           ( 1 + 2 sin(t) - 3 cos(t) )


<=>

 r - 2r sin(t) + 3 r cos(t) = 5 and r + 2r sin(t) - 3 r cos(t) = -5

<=>

 r = 5 + 2 r sin(t) - 3 r cos(t) and r = -( 5 + 2 r sin(t) - 3 r cos(t))

<=>

 r2 = ( 5 + 2 r sin(t) - 3 r cos(t))2

<=>

 x2 + y2 = (5 + 2 y - 3 x)2

<=>

 8x2 - 12 xy + 3 y2 - 30 x + 20 y + 25 = 0
For the line L, the transformation is easy.
 

L has equation  2 r cos(t) + r sin(t) = 2

<=>      2 x + y = 2
To calculate the intersection points in cartesian coordinates, we solve the system
 
 8x2 - 12 xy + 3 y2 - 30 x + 20 y + 25 = 0

 2 x + y = 2
With simple algebra we find
 
      ( 1.1217 ; -0.24347)  and (1.560 ; -1.120 )
It is easy to verify that these points are the same points as above.

Rotating and the polar equation.

Say alpha is a constant value. Take the curves
 
        c with equation F(r,t)=0                (1)
and
        c' with equation F(r, t - alpha)=0      (2)

Now we have

P(ro,to) is a solution of (1)  <=> P(ro,to + alpha) is a solution of (2)
This means that we obtain the curve c' by rotating the curve c by an angle alpha (center of rotation is O).

Example 1:

If we rotate the circle r = 4cos(t) by an angle of pi/2 radians, the new circle has equation

 
        r = 4 cos(t - pi/2)
<=>     r = 4 sin(t)
We rotate the curve c with equation F(r,t)=0 by an angle alpha. The new curve has equation F(r, t - alpha)=0.

Example 2:

We rotate the curve c with equation r = sin(2t) by an angle of pi radians. The new curve c' has equation

 
  r =  sin(2(t-pi))  <=> r = sin(2t - 2pi) <=> r = sin(2t)
We find that c and c' have the same equation. This means that c is invariant for a rotation over pi radians.

Asymptotes in polar coordinates

Visualization of dr/dt

 
    
Suppose a curve c has polar equation r = f(t).
At a variable point P of the curve, we draw a tangent line and on that line we denote the axis b in the direction of increasing t-values.
This defines the angle n and the angle a.
Now rotate axis u by pi/2 radians. This gives axis s.
The intersection point of b and s is T.
Denote N on s such that PN is perpendicular to TP. (see figure)
B is the unit vector with abscissa 1 relative to the axis b.
S is the unit vector with abscissa 1 relative to the axis s.
Now we have (vectors in bold)

Asymptotes

Say r = f(t) is a polar equation of a curve c.
First we write this equation as 1/r = g(t).
If to is a solution of g(t)=0, then to gives the direction of the asymptote.
From above, we have :
 
        1      d  1
        -  =  -- (-) = g'(t)  <=> m = 1/ g'(t)
        m     dt  r

So,
        mo = 1/ g'(to)
From mo we denote point T and then we draw the asymptote.
 
    
The equation of the asymptote is
 
                mo
        r = ---------------
            cos(t-(to+pi/2))

                 mo
      <=> r = ----------
              sin(t -to)


To calculate an asymptotes of the curve c with polar equation r = f(t), we take four steps:
  • Write g(t) = 1/f(t).
  • Calculate to = a solution of g(t)=0.
    to gives the direction of the asymptote.
  • Calculate mo = 1/ g'(to).
  • The asymptote is the line through point (mo,to+pi/2) and with direction to. Its equation is
     
                     mo
              r = -----------
                  sin(t - to)
    

3 Example:

Take the curve c with equation r = 3/(1+2cos(t)).
Then g(t) = (1+2cos(t))/3 and g'(t) = -2sin(t)/3
g(t) = 0 for to = 2pi/3 and t1 = -2pi/3
These values are the directions of the asymptotes.
The corresponding values of m are
mo = - sqrt(3) and m1 = sqrt(3).
The asymptotes are

 
                - sqrt(3)                        sqrt(3)
        r = ---------------  and        r = ---------------
             sin(t - 2pi/3)                  sin(t + 2pi/3)
In a graph this gives a hyperbola and his two asymptotes.
 
            
Exercise: Find the asymptote of the curve r = 2 tan(t) . sin(t)

Area with polar coordinates

General formula

 
        
A curve has polar equation r = f(t). Consider the colored part bounded by OAB. One can prove that the area of this part is equal to
 
    t2
  /
  |   1
  |   - r2 dt
  |   2
  /
  t1

Area calculations

A cardioid has polar equation r = 2 a (1 + cos(t)). On the graph is a = 1/2.
 
        
Since the curve is symmetrical relative to the polar axis, we first calculate the area of the upper half. The area of this part is:
 
    pi
  /
  |   1
  |   -  4 a2 (1 + cos(t))2 dt
  |   2
  /
  0

         pi
       /
       |
 =2a2 |  (1 + 2 cos(t) + cos2(t)) dt
       |
       /
       0

              now:  cos2(t) = (1/2) (1 + cos(2t))

                                                     pi
 = 2 a2 [ t + 2 sin(t) + (1/2)( t + (1/2) sin(2t) ]
                                                     0

 =  2 a2  (3/2) pi = 3 pi a2
The area of the whole cardioid is 6 pi a2

Exercise 1:
The polar equation r = sin(3t) gives a three-bladed curve. Plot the curve. Find the area of 1 blade.

 
        

Exercise 2:
Calculate the area of the part of the circle r = 4 cos(t) located outside the circle r = 1.5

 
        

Exercise 3:
Calculate the area of the region bounded by the smallest loop of the curve r = 1 + 2 cos(t).

 
        

Arc length in polar coordinates

 
        
A curve has a polar equation r=f(t). Consider the arc AB.
Let r' = dr/dt.
One can prove that the arc length of this part is equal to
 
    t2
  /
  |
  |   sqrt( r2 + r'2) dt
  |
  /
  t1

Length of a cardioid

We calculate the length of the cardioid r = 1+cos(t).
 
    

   r' = - sin(t)

   r2 + r'2 = 1 + 2 cos(t) + cos2(t) + sin2(t)

        = 2 (1 + cos(t))

        = 4 cos2(t/2)

   The length a half cardioid is then

    pi
  /
  |                                   pi
  |   2 cos(t/2) dt  = 2 [2 sin(t/2)]      = 4
  |                                   0
  /
  0

The total length is 8.
The length of r = a.(1+cos(t)) = 8 a






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